已知圆C:x^2+(y-1)^2=5,直线:mx-y+1-m=0

(1) L: mx-y+1-m=0--->m(x-1)=y-1--->L恒过定点(1,1)
而 1²+(1-1)²＜5，即该定点在圆C内
∴对m∈R,直线L与圆C总有两个不同的交点

(2) --->d²(C,L) = r²-(|AB|/2)² = 5-17/4 = 3/4
= (-1+1-m)²/(m²+1)
--->4m²=3m²+1--->m²=1--->m=±1--->L的倾斜角=45°或135°

（3）圆C:x^2+(y-1)^2=5-----------------------1
l：mx-y+1-m=0---------------------------2
联立1，2得
(1+m^2)x^2-2m^2x+m^2-5=0
令X=(x1+x2)/2=m^2/(1+m^2)---------------3
Y=(y1+y2)/2=(m^2-m+1)/(1+m^2)-----------4
X=1-1/(1+m^2)
Y=1-m/(1+m^2)
(X-1+Y-1)^2=（1+m）^2/(1+m^2)
=1+2m/(1+m^2)
=1+2(1-Y)
(X-1+Y-1)^2=1+2(1-Y)
（x-1/2)^2+ (y-1)^2=1/4